我试图更好地学习PHP我的自定义框架(学习目的),我想知道如何实现带别名的命名空间(他们对我来说是新手)。我的#34;框架"结构看起来像:
app/
controllers/
core/
functions/
languages/
models/
vendor/
views/
.htaccess
composer.json
composer.lock
config.php
database.php
filters.php
index.php
init.php
routes.php
style/
upload/
.htaccess
index.php
robots.txt
在app / core文件夹中我有文件:
App.php
Controller.php
Filter.php
Functions.php
Language.php
Route.php
index.php
我的init.php文件如下:
<?php
require "vendor/autoload.php";
require "database.php";
require "core/App.php";
require "core/Filter.php";
require "core/Route.php";
require "functions/Common.php";
require "functions/Validator.php";
require "functions/Custom.php";
require "core/Language.php";
require "core/Functions.php";
require "core/Controller.php";
来自controllers /文件夹的文件扩展了Controller.php,core / Functions.php扩展了core / Custom.php,扩展了core / Common.php 我如何实现命名空间和alisaing,这样我可以使我的框架更容易维护,而不需要使用完整的命名空间加载每个类?提前致谢
修改: 好的我已经在app / controllers /中使用此代码
创建了一个文件controller.php<?php
use App\Core\Controller as BaseController;
class Controller extends BaseController {
}
在我的App.php中,我添加了以下功能:
private function applyAliases() {
foreach(self::$config['aliases'] as $file => $alias) {
class_alias($file, $alias);
}
}
所以在我的config.php文件中我添加了:
/* Aliases */
"aliases" => [
"App\Core\Language" => "Language",
"App\Core\Filter" => "Filter",
"App\Core\Route" => "Route"
]
另外,在Functions.php中加载函数时我添加了另一个class_alias:
public function loadFunction(string $file) {
if(!file_exists(App::$config['paths']['functions'].$file.".php")) {
throw new Exception("File ".App::$config['paths']['functions'].$file.".php doesn't exists");
}
require App::$config['paths']['functions'].$file.".php";
class_alias("App\Functions\\".$file, $file);
return new $file();
}
这会是正确的方法吗?